pat00-自测4. Have Fun with Numbers (20)

00-自测4. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

---

提交代码

#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <cmath>
#include <iostream>
using namespace std;
int ti[10],num1[25],num2[25];
char num[25];
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    scanf("%s",num);

    //cout<<num<<endl;

    int i;
    for(i=0;i<strlen(num);i++){
        num1[i]=num[i]-'0';
        ti[num1[i]]++;
        num2[i]=2*num1[i];
    }
    int k=0,t;
    for(i=strlen(num)-1;i>=0;i--){
        t=num2[i]+k;
        num2[i]=t%10;
        ti[num2[i]]--;
        k=t/10;
    }
    bool can=false;
    if(!k){
        for(i=0;i<=9;i++){
            if(ti[i]){
                break;
            }
        }
        if(i==10){
            can=true;
        }
    }
    if(can){
        printf("Yes
");
    }
    else{
        printf("No
");
    }
    if(k){
        printf("%d",k);
    }
    for(i=0;i<strlen(num);i++){
        printf("%d",num2[i]);
    }
    printf("
");
    return 0;
}